Peter Freyd's and John Kennison's examples definitively settled Mike Barr's question about canonical subobjects that compose. But I had started thinking about it and had what I thought would be a nice example. The category of sets has canonical quotients (equivalence classes) but they don't compose. I think there is no choice that do, but so far I haven't been able to prove or disprove this. Anybody? Bob [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Robert Pare hat am 12.09.10 geschrieben:
Peter Freyd's and John Kennison's examples definitively settled Mike Barr's question about canonical subobjects that compose. But I had started thinking about it and had what I thought would be a nice example. The category of sets has canonical quotients (equivalence classes) but they don't compose. I think there is no choice that do, but so far I haven't been able to prove or disprove this. Anybody?
There is. First consider the full subcategory of partitions; that is, sets whose elements happen to be non-empty, pairwise disjoint sets. It has an obvious choice of quotient maps that does the trick, namely those maps for which each element of the target is the union of its fibre. For the remaining sets as sources, additionally choose the identity in case of the trivial quotient, the canonical map otherwise. Thorsten [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
[note from moderator: post delayed from Monday by a transmission error] Hi Bob, how about this: using a sufficiently strong version of the axiom of choice, assume a well-ordering on the class of all sets (the "global well-ordering"). (Someone will be able to tell me whether this is strictly stronger than the ordinary axiom of choice, or equivalent to it. In any case, assuming set theory is consistent in the first place, and sufficiently large cardinals exist, there certainly exist models of set theory with such a property). Given any equivalence relation ~ on a set X, say that an element x of X is a "canonical representative" of ~ if x is the (unique) least element in its equivalence class under the global well-ordering. Let X/~ be the set of canonical representatives, and define the map X -> X/~ to pick out the canonical representative of each class. I think this gives canonical quotients on (this version of) Set. -- Peter Robert Pare wrote:
Peter Freyd's and John Kennison's examples definitively settled Mike Barr's question about canonical subobjects that compose. But I had started thinking about it and had what I thought would be a nice example. The category of sets has canonical quotients (equivalence classes) but they don't compose. I think there is no choice that do, but so far I haven't been able to prove or disprove this. Anybody?
Bob
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
But that's only an equivalent category; everyone knows that can be done. Even easier is the dual category of complete atomic boolean algebras, which has canonical subobjects. On Sun, 12 Sep 2010, Thorsten Palm wrote:
Robert Pare hat am 12.09.10 geschrieben:
Peter Freyd's and John Kennison's examples definitively settled Mike Barr's question about canonical subobjects that compose. But I had started thinking about it and had what I thought would be a nice example. The category of sets has canonical quotients (equivalence classes) but they don't compose. I think there is no choice that do, but so far I haven't been able to prove or disprove this. Anybody?
There is. First consider the full subcategory of partitions; that is, sets whose elements happen to be non-empty, pairwise disjoint sets. It has an obvious choice of quotient maps that does the trick, namely those maps for which each element of the target is the union of its fibre. For the remaining sets as sources, additionally choose the identity in case of the trivial quotient, the canonical map otherwise.
Thorsten
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Michael Barr hat am 13.09.10 geschrieben:
But that's only an equivalent category; everyone knows that can be done.
Yes, as far as that category of partitions is concerned. But note my last sentence:
On Sun, 12 Sep 2010, Thorsten Palm wrote:
For the remaining sets as sources, additionally choose the identity in case of the trivial quotient, the canonical map otherwise.
Perhaps I should have put the emphasis differently: for a proper quotient of a non-partition, choose the canonical map (its range is a partition); for a quotient of a partition, use the union construction. Of course the case distinction is somewhat unsatisfactory, and so I was trying to hint that there would be a nicer solution if all sets were partitions in the first place. It might be worth mentioning that a similar idea serves to equip the very category of sets with a (binary-)product functor to make it strictly monoidal. Thorsten [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
participants (4)
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Michael Barr -
pare＠mathstat.dal.ca -
selinger＠mathstat.dal.ca -
Thorsten Palm